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2 years ago

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The gravitational force between two asteroids is 6. 2 × 108 N. Asteroid Y has three times the mass of asteroid Z. If the distanc

e between the asteroids is 2100 kilometers, what is the mass of asteroid Y? 3. 7 × 1015 kg 1. 1 × 1016 kg 1. 4 × 1031 kg 4. 1 × 1031 kg.

1 answer:

Sliva [168]2 years ago

7 0

Answer:Let m = mass of asteroid y.Because asteroid y has three times the mass of asteroid z, the mass of asteroid z is m/3.Given:F = 6.2x10⁸ Nd = 2100 km = 2.1x10⁶ mNote thatG = 6.67408x10⁻¹¹ m³/(kg-s²)The gravitational force between the asteroids isF = (G*m*(m/3))/d² = (Gm²)/(3d²)orm² = (3Fd²)/G = [(3*(6.2x10⁸ N)*(2.1x10⁶ m)²]/(6.67408x10⁻¹¹ m³/(kg-s²)) = 1.229x10³² kg²m = 1.1086x10¹⁶ kg = 1.1x10¹⁶ kg (approx)Answer: 1.1x10¹⁶ kg

Explanation:

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Emmy is standing on a moving sidewalk that moves at +2 m/s. Suddenly, she realizes she might miss her flight, so begins to speed

likoan [24]

Answer:

Refer to the attachment for the diagram.

3.53 m/s.

Explanation:

Acceleration is the first derivative of velocity relative to time. In other words, the acceleration is the same as the slope (gradient) of the velocity-time graph. Let $t$ represents the time in seconds and $v$ the speed in meters-per-second.

For $0 < x \le 1$ :

Initial value of $v$ : $\rm 2\;m\cdot s^{-1}$ at $t = 0$ ; Hence the point on the segment: $(0, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 2\; m\cdot s^{-2}$ .
Find the equation of this segment in slope-point form: $v - 2 = 2 (t - 0) \implies v = 2t + 2, \quad 0 < t \le 1$ .

Similarly, for $1 < x \le 2$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 1$ : $t = 2t + 2 = 4$ ; Hence the point on the segment: $(1, 4)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 1\; m\cdot s^{-2}$ .

Find the equation of this segment in slope-point form: $v - 4 = (t - 1) \implies v = t + 3 \quad 1 < t \le 2$ .

For $2 < x \le 3$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 2$ : $t = t + 3 = 5$ ; Hence the point on the segment: $(2, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . There's no acceleration. In other words, the velocity is constant.
Find the equation of this segment in slope-point form: $v - 5 = 0 (t - 2) \implies v = 5 \quad 2 < t \le 3$ .

For $3 < x \le 4$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 3$ : $t = 5$ ; Hence the point on the segment: $(3, 5)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm -3\; m\cdot s^{-2}$ . In other words, the velocity is decreasing.
Find the equation of this segment in slope-point form: $v - 5 = -3 (t - 3) \implies v = -3t + 14 \quad 3 < t \le 4$ .

For $4 < x \le 5$ :

Initial value of $v$ is the same as the final value of $v$ in the previous equation at $t = 4$ : $t = -3t + 14$ ; Hence the point on the segment: $(4, 2)$ .
Slope of the velocity-time graph is the same as acceleration during that period of time: $\rm 0\; m\cdot s^{-2}$ . In other words, the velocity is once again constant.
Find the equation of this segment in slope-point form: $v - 2 = 0 (t - 4) \implies v = 2\quad 4 < t \le 5$ .

$t = \rm 3.49\;s$ is in the interval $3 < x \le 4$ . Apply the equation for that interval: $v = -3t +14 = \rm 3.53\; m \cdot s^{-1}$ .

4 0

3 years ago

. A chemistry student’s height is measured at 68.5 inches. How tall is the student in cm?

tatiyna

1 inch = 2.54 centimeters

All we need to do is multiply.

68.5 * 2.54 = 173.99cm

Best of Luck!

3 0

3 years ago

Read 2 more answers

A LED light source contains a 0.5-Watts GaAs (Eg =1.43 eV) LED. Assuming that 0.12% of the electric energy is converted to emiss

Ray Of Light [21]

Answer:

Explanation:

energy emitted by source per second = .5 J

Eg = 1.43 eV .

Energy converted into radiation = .5 x .12 = .06 J

energy of one photon = 1.43 eV

= 1.43 x 1.6 x 10⁻¹⁹ J

= 2.288 x 10⁻¹⁹ J .

no of photons generated = .06 / 2.288 x 10⁻¹⁹

= 2.6223 x 10¹⁷

wavelength of photon λ = 1275 / 1.43 nm

= 891.6 nm .

momentum of photon = h / λ ; h is plank's constant

= 6.6 x 10⁻³⁴ / 891.6 x 10⁻⁹

= .0074 x 10⁻²⁵ J.s

Total momentum of all the photons generated

= .0074 x 10⁻²⁵ x 2.6223 x 10¹⁷

= .0194 x 10⁻⁸ Js

b ) spectral width in terms of wavelength = 30 nm

frequency width = ?

n = c / λ , n is frequency , c is velocity of light and λ is wavelength

differentiating both sides

dn = c x dλ / λ²

given dλ = 30 nm

λ = 891.6 nm

dn = 3 x 10⁸ x 30 x 10⁻⁹ / ( 891.6 x 10⁻⁹ )²

= 11.3 x 10¹² Hz .

c )

10 nW = 10 x 10⁻⁹ W

= 10⁻⁸ W .

energy of 50 dB

50 dB = 5 B

I / I₀ = 10⁵ ; decibel scale is logarithmic , I is energy of sound having dB = 50 and I₀ = 10⁻¹² W /s

I = I₀ x 10⁵

= 10⁻¹² x 10⁵

= 10⁻⁷ W

= 10 x 10⁻⁸ W

power required

= 10⁻⁸ + 10 x 10⁻⁸ W

= 11 x 10⁻⁸ W.

5 0

3 years ago

Please help me out i'm so depressed and such a failure

aksik [14]

Answer:

I reckon towards b. Let me know if im right

4 0

3 years ago

How much work is done when a porcelain vase with mass of 2.5 kg is lifted up 3 m to a shelf? (HINT- Weight can be used as Force

alexdok [17]

Work = force x distance
F= 2.5
D= 3
Work = 2.5 x 3 =7.5
Work = 7.5 J
J=Jules (Jules is the unit uses to calculate work)

8 0

3 years ago